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JEE Main 2010
Exam
JEE Main
Year
2010
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2010 · Physics · Thermal Properties of Matter

Two conductors have the same resistance at 0^ C but their...

Two conductors have the same resistance at 0∘C0^{\circ} \mathrm{C} but their temperature coefficients of resistance are α1\alpha_1 and α2\alpha_2. The respective temperature coefficients of their series and parallel combinations are nearly

Options

  1. Aα1+α22,α1+α2\frac{\alpha_1+\alpha_2}{2}, \alpha_1+\alpha_2
  2. Bα1+α2,α1+α22\alpha_1+\alpha_2, \frac{\alpha_1+\alpha_2}{2}
  3. Cα1+α2,α1α2α1+α2\alpha_1+\alpha_2, \frac{\alpha_1 \alpha_2}{\alpha_1+\alpha_2}
  4. Dα1+α22,α1+α22\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_1+\alpha_2}{2}

Correct answer

D. α1+α22,α1+α22\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_1+\alpha_2}{2}

Step-by-step solution

Let R0R_0 be the initial resistance of both conductors ∴\therefore \quad At temperature θ\theta their resistance will be, R1=R0(1+α1θ)R_1=R_0\left(1+\alpha_1 \theta\right) and R2=R0(1+α2θ)R_2=R_0\left(1+\alpha_2 \theta\right) for, series combination, Rs=R1+R2Rs0(1+αsθ)\mathrm{R}_{\mathrm{s}}=\mathrm{R}_1+\mathrm{R}_2 R_{s 0}\left(1+\alpha_s \theta\right)=R_0\left(1+\alpha_1 \theta\right)+R0(1+α2θ)+R_0\left(1+\alpha_2 \theta\right) where Rs0=R0+R0=2R0\mathrm{R}_{\mathrm{s} 0}=\mathrm{R}_0+\mathrm{R}_0=2 \mathrm{R}_0 \therefore \quad 2 R0(1+αsθ)=2R0+R0\mathrm{R}_0\left(1+\alpha_s \theta\right)=2 \mathrm{R}_0+\mathrm{R}_0 \theta\left(\alpha_1+\alpha_2\right)oror\quad \alpha_s=\frac{\alpha_1+\alpha_2}{2}forparallelcombination,Rp=for parallel combination, R_p=\frac{R_1 R_2}{R_1+R_2}where,where,R_{p 0}=\frac{R_0 R_0}{R_0+R_0}=\frac{R_0}{2} \therefore \quad \frac{\mathrm{R}_0}{2}(1+αpθ)=R02(1+α1θ+α2θ+α1α2θ)R0(2+α1θ+α2θ)\left(1+\alpha_{\mathrm{p}} \theta\right)=\frac{\mathrm{R}_0^2\left(1+\alpha_1 \theta+\alpha_2 \theta+\alpha_1 \alpha_2 \theta\right)}{\mathrm{R}_0\left(2+\alpha_1 \theta+\alpha_2 \theta\right)} as α1\alpha_1 and α2\alpha_2 are small quantities ∴α1α2\therefore \quad \alpha_1 \alpha_2 is negligible or αp=α1+α22+(α1+α2)θ=α1+α22[1−(α1+α2)θ]\quad \alpha_{\mathrm{p}}=\frac{\alpha_1+\alpha_2}{2+\left(\alpha_1+\alpha_2\right) \theta}=\frac{\alpha_1+\alpha_2}{2}\left[1-\left(\alpha_1+\alpha_2\right) \theta\right] as (α1+α2)2\left(\alpha_1+\alpha_2\right)^2 is negligible ∴αp=α1+α22\therefore \quad \alpha_p=\frac{\alpha_1+\alpha_2}{2}
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